A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm `s^(-1)` in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of `10^(-3) T cm^(-1)` along the negative x-direction (that is it increases by `10^(-3) T cm^(-1 )`as one moves in the negative x-direction), and it is decreasing in time at the rate of `10^(-3) T s^(-1)`. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 `mOmega`.

Explanation

Rate of change of flux due to explicit time variation in B
`= 144 xx 10^(-4) m^2 xx 10^(-3) T s^(-1)`
`= 1.44 xx 10^(-5) Wbs^(-1)`
Rate of change of flux due to motion of the loop in a non-uniform B
`=144 xx 10^(-4) m^2 xx 10^(-3) T cm^(-1) xx 8 cm s^(-1)`
`= 11.52 xx 10^(5) Wb s^(-1)`
The two effects add up since both cause a decrease in flux along the positive z-direction. Therefore, induced emf `= 12.96 ×x 10^(-5) V`, induced current = `2.88 x× 10^(-2)` A. The direction of induced current is such as to increase the flux through the loop along positive z-direction. If for the observer the loop moves to the right, the current will be seen to be anti-clockwise. A proper proof of the procedure above is as follows:
`Phi (t) = int_0^a aB (x,t)dx`
`(dPhi)/(dt) = a int_(0)^a dx (dB(x,t))/(dt)`
using,
`(dB)/(dt) = (deltaB)/(deltat) + (deltaB)/(deltax) (dx)/(dt)`
`=[(deltaB)/(deltat) + v(deltaB)/(deltax)]`
we get,
`(dPhi)/(dt) = a int_(0)^a dx [(deltaB(x,t))/(deltat) + v(deltaB(x,t))/(deltax)]`
`=A [(deltaB)/(deltat) + v(deltaB)/(deltax)]`
where A= `a^2`
The last step follows because `((deltaB)/(deltat)) , ((deltaB)/(deltax))` and v are given to be
constants in the problem. Even if you do not understand this formal proof (which requires good familiarity with calculus), you will still appreciate that flux change can occur both due to the motion of the loop as well as time variations in the magnetic field.